Precalculus Examples

$- x^{2} + 3 y^{2} - 4 x + 6 y = 28$

Step 1

Complete the square for $- x^{2} - 4 x$ .

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Step 1.1

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = - 1$

$b = - 4$

$c = 0$

Step 1.2

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Step 1.3

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

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Step 1.3.1

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{- 4}{2 \cdot - 1}$

Step 1.3.2

Simplify the right side.

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Step 1.3.2.1

Cancel the common factor of $- 4$ and $2$ .

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Step 1.3.2.1.1

Factor $2$ out of $- 4$ .

$d = \frac{2 \cdot - 2}{2 \cdot - 1}$

Step 1.3.2.1.2

Move the negative one from the denominator of $\frac{- 2}{- 1}$ .

$d = - 1 \cdot - 2$

Step 1.3.2.2

Rewrite $- 1 \cdot - 2$ as $- - 2$ .

$d = - - 2$

Step 1.3.2.3

Multiply $- 1$ by $- 2$ .

$d = 2$

Step 1.4

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

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Step 1.4.1

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{{(- 4)}^{2}}{4 \cdot - 1}$

Step 1.4.2

Simplify the right side.

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Step 1.4.2.1

Simplify each term.

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Step 1.4.2.1.1

Cancel the common factor of ${(- 4)}^{2}$ and $4$ .

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Step 1.4.2.1.1.1

Rewrite $- 4$ as $- 1 (4)$ .

$e = 0 - \frac{{(- 1 (4))}^{2}}{4 \cdot - 1}$

Step 1.4.2.1.1.2

Apply the product rule to $- 1 (4)$ .

$e = 0 - \frac{{(- 1)}^{2} \cdot 4^{2}}{4 \cdot - 1}$

Step 1.4.2.1.1.3

Raise $- 1$ to the power of $2$ .

$e = 0 - \frac{1 \cdot 4^{2}}{4 \cdot - 1}$

Step 1.4.2.1.1.4

Multiply $4^{2}$ by $1$ .

$e = 0 - \frac{4^{2}}{4 \cdot - 1}$

Step 1.4.2.1.1.5

Factor $4$ out of $4^{2}$ .

$e = 0 - \frac{4 \cdot 4}{4 \cdot - 1}$

Step 1.4.2.1.1.6

Move the negative one from the denominator of $\frac{4}{- 1}$ .

$e = 0 - (- 1 \cdot 4)$

Step 1.4.2.1.2

Multiply $- (- 1 \cdot 4)$ .

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Step 1.4.2.1.2.1

Multiply $- 1$ by $4$ .

$e = 0 - - 4$

Step 1.4.2.1.2.2

Multiply $- 1$ by $- 4$ .

$e = 0 + 4$

Step 1.4.2.2

Add $0$ and $4$ .

$e = 4$

Step 1.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(x + 2)}^{2} + 4$ .

$- {(x + 2)}^{2} + 4$

Step 2

Substitute $- {(x + 2)}^{2} + 4$ for $- x^{2} - 4 x$ in the equation $- x^{2} + 3 y^{2} - 4 x + 6 y = 28$ .

$- {(x + 2)}^{2} + 4 + 3 y^{2} + 6 y = 28$

Step 3

Move $4$ to the right side of the equation by adding $4$ to both sides.

$- {(x + 2)}^{2} + 3 y^{2} + 6 y = 28 - 4$

Step 4

Complete the square for $3 y^{2} + 6 y$ .

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Step 4.1

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 3$

$b = 6$

$c = 0$

Step 4.2

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Step 4.3

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

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Step 4.3.1

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{6}{2 \cdot 3}$

Step 4.3.2

Simplify the right side.

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Step 4.3.2.1

Cancel the common factor of $6$ and $2$ .

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Step 4.3.2.1.1

Factor $2$ out of $6$ .

$d = \frac{2 \cdot 3}{2 \cdot 3}$

Step 4.3.2.1.2

Cancel the common factors.

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Step 4.3.2.1.2.1

Factor $2$ out of $2 \cdot 3$ .

$d = \frac{2 \cdot 3}{2 (3)}$

Step 4.3.2.1.2.2

Cancel the common factor.

$d = \frac{2 \cdot 3}{2 \cdot 3}$

Step 4.3.2.1.2.3

Rewrite the expression.

$d = \frac{3}{3}$

Step 4.3.2.2

Cancel the common factor of $3$ .

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Step 4.3.2.2.1

Cancel the common factor.

$d = \frac{3}{3}$

Step 4.3.2.2.2

Rewrite the expression.

$d = 1$

Step 4.4

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

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Step 4.4.1

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{6^{2}}{4 \cdot 3}$

Step 4.4.2

Simplify the right side.

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Step 4.4.2.1

Simplify each term.

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Step 4.4.2.1.1

Raise $6$ to the power of $2$ .

$e = 0 - \frac{36}{4 \cdot 3}$

Step 4.4.2.1.2

Multiply $4$ by $3$ .

$e = 0 - \frac{36}{12}$

Step 4.4.2.1.3

Divide $36$ by $12$ .

$e = 0 - 1 \cdot 3$

Step 4.4.2.1.4

Multiply $- 1$ by $3$ .

$e = 0 - 3$

Step 4.4.2.2

Subtract $3$ from $0$ .

$e = - 3$

Step 4.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $3 {(y + 1)}^{2} - 3$ .

$3 {(y + 1)}^{2} - 3$

Step 5

Substitute $3 {(y + 1)}^{2} - 3$ for $3 y^{2} + 6 y$ in the equation $- x^{2} + 3 y^{2} - 4 x + 6 y = 28$ .

$- {(x + 2)}^{2} + 3 {(y + 1)}^{2} - 3 = 28 - 4$

Step 6

Move $- 3$ to the right side of the equation by adding $3$ to both sides.

$- {(x + 2)}^{2} + 3 {(y + 1)}^{2} = 28 - 4 + 3$

Step 7

Simplify $28 - 4 + 3$ .

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Step 7.1

Subtract $4$ from $28$ .

$- {(x + 2)}^{2} + 3 {(y + 1)}^{2} = 24 + 3$

Step 7.2

Add $24$ and $3$ .

$- {(x + 2)}^{2} + 3 {(y + 1)}^{2} = 27$

Step 8

Divide each term by $27$ to make the right side equal to one.

$- \frac{{(x + 2)}^{2}}{27} + \frac{3 {(y + 1)}^{2}}{27} = \frac{27}{27}$

Step 9

Simplify each term in the equation in order to set the right side equal to $1$ . The standard form of an ellipse or hyperbola requires the right side of the equation be $1$ .

$\frac{{(y + 1)}^{2}}{9} - \frac{{(x + 2)}^{2}}{27} = 1$